Let \(p\) be an odd prime. Show that \(\sum_{\beta \in \mathbb{Z}_p^*} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^*} \beta = 0\).

== Attempt ==

Given \(p\) is prime, \(\mathbb{Z}_p = \mathbb{Z}_p^*\), which means all integers from \(1\) to \((p - 1)\) are in \(\mathbb{Z}_p^*\). Furthermore, every element has a multiplicative inverse, and is itself a multiplicative inverse of some element. Thus,

\(\displaystyle\sum_{\beta \in \mathbb{Z}_p^*} \beta^{-1} = \sum_{\beta \in \mathbb{Z}_p^*} \beta = 1 + \dots + (p-1) = \frac{p(p-1)}{2} \)

Given \(p\) is an odd prime, \(2 \mid (p-1)\). Therefore,

\(\displaystyle \lbrack \frac{p(p-1)}{2} \rbrack_p = \lbrack p \rbrack_p = \lbrack 0 \rbrack_p \)

\(\Box\)